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Bạn kiểm tra lại đề. Theo mình
\(H=5\left(\sqrt{2+\sqrt{3}}-\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
Đặt \(A=\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}{\sqrt{2}\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}+1-\sqrt{3}+1}=\frac{2\sqrt{3}}{2}=\sqrt{3}\)
Vậy \(A=\sqrt{3}\)